Exams Joint Entrance Examination Mathematics Sequences and Series Question 39 Question 39 of 519 Medium MCQ
Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :
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S 3 = 3 a + 3 d = 54 ⇒ a + d = 18 S 20 = 10 ( 2 a + 19 d ) ⇒ 10 ( 36 + 17 d ) ⇒ 1600 < 10 ( 36 + 17 d ) < 1800 ⇒ 160 < 36 + 17 d < 180 ⇒ 124 < 17 d < 144 ⇒ 7 5 17 < d < 8 8 17 ⇒ d = 8 ⇒ a = 10 ⇒ a 11 = 10 + 10 × 8 = 90 \begin{gathered} {S}_{3}=3a+3d=54 \\ \Rightarrow a+d=18 \\ {S}_{20}=10(2a+19d) \\ \Rightarrow 10(36+17d) \\ \Rightarrow 1600<10(36+17d)<1800 \\ \Rightarrow 160<36+17d<180 \\ \Rightarrow 124<17d<144 \\ \Rightarrow 7\frac{5}{17}<d<8\frac{8}{17} \\ \\ \Rightarrow d=8\Rightarrow a=10 \\ \Rightarrow {a}_{11}=10+10\times 8=90 \end{gathered} S 3 = 3 a + 3 d = 54 ⇒ a + d = 18 S 20 = 10 ( 2 a + 19 d ) ⇒ 10 ( 36 + 17 d ) ⇒ 1600 < 10 ( 36 + 17 d ) < 1800 ⇒ 160 < 36 + 17 d < 180 ⇒ 124 < 17 d < 144 ⇒ 7 17 5 < d < 8 17 8 ⇒ d = 8 ⇒ a = 10 ⇒ a 11 = 10 + 10 × 8 = 90 Answer the question to unlock